How are $0.50 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{CO}_{3}$ and $0.50 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ different?

Solution:

Molar mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}=(2 \times 23)+12.00+(3 \times 16)=106 \mathrm{~g} \mathrm{~mol}^{-1}$

Now, 1 mole of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ means $106 \mathrm{~g}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$.

$\therefore 0.5 \mathrm{~mol}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3} \quad=\frac{106 \mathrm{~g}}{1 \mathrm{~mole}} \times 0.5 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{CO}_{3}$

$=53 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}$

$\Rightarrow 0.50 \mathrm{M}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}=0.50 \mathrm{~mol} / \mathrm{L} \mathrm{Na}_{2} \mathrm{CO}_{3}$

Hence, $0.50 \mathrm{~mol}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is present in $1 \mathrm{~L}$ of water or $53 \mathrm{~g}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is present in $1 \mathrm{~L}$ of water.